Projectile on Inclined Plane — Maximum Range

The question

For a given launch speed $u$ on an inclined plane of angle $\beta$, at what angle $\alpha$ should the body be projected to achieve maximum range up the slope? And what is that maximum range?

Starting point

The range up the inclined plane is:

$R = \frac{2u^2\sin(\alpha-\beta)\cos\alpha}{g\cos^2\beta}$

To maximise $R$, we need to maximise $\sin(\alpha-\beta)\cos\alpha$ with respect to $\alpha$.

Finding the optimal angle

Use the product-to-sum identity:

$2\sin(\alpha-\beta)\cos\alpha = \sin(\alpha - \beta + \alpha) + \sin(\alpha - \beta - \alpha)$

$= \sin(2\alpha - \beta) + \sin(-\beta)$

$= \sin(2\alpha - \beta) - \sin\beta$

So:

$R = \frac{u^2}{g\cos^2\beta}\left[\sin(2\alpha - \beta) - \sin\beta\right]$

$R$ is maximum when $\sin(2\alpha - \beta)$ is maximum, i.e. when $\sin(2\alpha - \beta) = 1$:

$2\alpha - \beta = 90°$

$\boxed{\alpha_{\text{opt}} = \frac{90° + \beta}{2} = 45° + \frac{\beta}{2}}$

The optimal angle is $45°$ plus half the slope angle.

Maximum range

Substituting $\sin(2\alpha - \beta) = 1$:

$R_{\max} = \frac{u^2}{g\cos^2\beta}(1 - \sin\beta)$

Since $\cos^2\beta = 1 - \sin^2\beta = (1-\sin\beta)(1+\sin\beta)$:

$R_{\max} = \frac{u^2}{g(1-\sin\beta)(1+\sin\beta)} \cdot (1-\sin\beta)$

$\boxed{R_{\max} = \frac{u^2}{g(1+\sin\beta)}}$

Verification with flat ground ($\beta = 0$)

$\alpha_{\text{opt}} = 45° + 0° = 45° \checkmark$

$R_{\max} = \frac{u^2}{g(1+0)} = \frac{u^2}{g} \checkmark$

Both results reduce correctly to the flat-ground case.

How slope angle affects maximum range

As $\beta$ increases (steeper slope):

• $\alpha_{\text{opt}}$ increases — you must aim higher to maximise range
• $R_{\max}$ decreases — you cannot throw as far up a steeper slope